Question

A dolphin jumps out of the water. The function h=-16t^2+26t models the height (in feet) of the dolphin after t seconds. After how many seconds is the dolphin at a height of 5 feet.

Answer 1

Answer:

The dolphin is five feet in the air after about 0.22 and 1.40 seconds.

Step-by-step explanation:

The height h (in feet) of the dolphin as it jumped out of the water after t seconds is given by the function:

$h(t)=-16t^2+26t$

We want to determine the time(s) when the dolphin is five feet in the air.

Since the dolphin is five feet in the air, h(t) = 5:

$5=-16t^2+26t$

Solve for t. Rearrange:

$16t^2-26t+5=0$

We can use the quadratic formula, given by:

$\displaystyle t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 16, b = -26, and c = 5.

Substitute:

$\displaystyle t=\frac{-(-26)\pm\sqrt{(-26)^2-4(16)(5)}}{2(16)}$

Evaluate:

$\displaystyle t=\frac{26\pm\sqrt{356}}{32}$

Simplify the square root:

$\sqrt{356}=\sqrt{2\cdot 2\cdot 89}=2\sqrt{89}$

Hence:

$\displaystyle t=\frac{26\pm2\sqrt{89}}{32}$

Simplify:

$\displaystyle t=\frac{13\pm\sqrt{89}}{16}$

Hence, our solutions are:

$\displaystyle t=\frac{13-\sqrt{89}}{16}\approx 0.22\text{ and } t=\frac{13+\sqrt{89}}{16}\approx1.40$

The dolphin is five feet in the air after about 0.22 and 1.40 seconds.